How can I determine the Jordan Form of a matrix? How can I go about proving that the characteristic polynomial, minimal polynomial, and the dim(eigenspace) is enough to determine the Jordan Form of a matrix for n<7?
 A: Let's give it a try:
$$A=\begin{pmatrix}\color{red}0&\color{red}1&\color{red}0&0&0&0&0\\
\color{red}0&\color{red}0&\color{red}1&0&0&0&0\\
\color{red}0&\color{red}0&\color{red}0&0&0&0&0\\
0&0&0&\color{red}0&\color{red}1&0&0\\
0&0&0&\color{red}0&\color{red}0&0&0\\
0&0&0&0&0&1&0\\
0&0&0&0&0&0&1\end{pmatrix}\;,\;\;\;
B=\begin{pmatrix}\color{blue}0&\color{blue}1&\color{blue}0&\color{blue}0&0&0&0&\\
\color{blue}0&\color{blue}0&\color{blue}1&\color{blue}0&0&0&0\\
\color{blue}0&\color{blue}0&\color{blue}0&\color{blue}1&0&0&0\\
\color{blue}0&\color{blue}0&\color{blue}0&\color{blue}0&0&0&0\\
0&0&0&0&\color{blue}0&0&0\\
0&0&0&0&0&1&0\\
0&0&0&0&0&0&1\end{pmatrix}$$
Clearly $\,A\nsim B\,$ since one has Jordan zero blocks of size $\,3,2\,$ resp. (in red) and the other one two Jordan zero blocks of size $\,4,1\,$ resp. (in blue) .
Both matrices have $\,x^5(x-1)^2\,$ as characteristic polynomial, both have $\,x^4(x-1)\,$ as minimal polynomial and both have zero eigenspace of dimension $\,5\,$ (check this!)
A: I realize this is 5 years old, but the only answer is wrong, given that the two matrices in DonAntonio's answer do not have matching minimal polynomials. So in case anyone comes across it, here is an answer:
If Dimension=7, then you could have two matrices such as
$$
A=\begin{pmatrix}
  0 & 1 \\
   & 0 & 1 \\
   &  & 0 \\
   & & & 0 & 1 \\
   & & & & 0 & 1 \\
   & & & & & 0 \\
   & & & & & & 0
\end{pmatrix}
\quad,\quad
B=\begin{pmatrix}
  0 & 1 \\
   & 0 & 1 \\
   &  & 0 \\
   & & & 0 & 1 \\
   & & & & 0 &  \\
   & & & & & 0 & 1 \\
   & & & & & & 0
\end{pmatrix}
$$
Notice that the eigenspace is just the kernel, and both matrices are rank 4, so both have kernels of dimension 3. Both have characteristic polynomial $x^7$, and both have minimal polynomial $x^3$. But let's take a moment to examine what failed.
If we examine the number of integer partitions of $7$, we have
7=3+3+1 and also 7=3+2+2. These are both integer partitions of 7 into 3 parts with largest part 3. If we associate each part with a Jordan block, the number of parts is the dimension of the Eigenspace, and the largest part in the decomposition is the degree of the minimal polynomial. However, if we look at all partitions of 6, we have
6=6
6=5+1
6=4+2
6=4+1+1
6=3+3
6=3+2+1
6=3+1+1+1
6=2+2+2
6=2+2+1+1
6=2+1+1+1+1
6=1+1+1+1+1+1

Notice that for $6$, or any number smaller than 6 (as all of those are implicitly in this list as well), there is no integer partition for which the largest part and number of parts is the same simultaneously. This is why knowing the dimension of the eigenspace (tells you the number of parts), minimal polynomial (degree tells you the largest part), and characteristic polynomial (degree tells you the dimension of the ambient space) is sufficient, but it is not sufficient for dimension 7.
Best,
--Kris
